Ôn tập toán 8

Áp dụng bđt \(\left|a\right|+\left| b\right|\ge\left|a+b\right|\) , dấu "=" xảy ra khi a,b cùng dấu.

a) Ta có \(C=\left|x-1\right|+\left|x-4\right|=\left|x-1\right|+\left|4-x\right|\ge\left|x-1+4-x\right|=3\)

Dấu "=" xảy ra khi \(1\le x\le4\)

Vậy Min C = 3 tại \(1\le x\le4\)

b) Ta có : \(D=\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{4}\right|\)

\(=\left(\left|-x-\frac{1}{2}\right|+\left|x+\frac{1}{4}\right|\right)+\left|x+\frac{1}{3}\right|\)

Áp dụng bđt trên , ta được \(\left|-x-\frac{1}{2}\right|+\left|x+\frac{1}{4}\right|\ge\left|-x-\frac{1}{2}+x+\frac{1}{4}\right|=\frac{1}{4}\)

Lại có \(\left|x+\frac{1}{3}\right|\ge0\)

\(\Rightarrow D\ge\frac{1}{4}+0=\frac{1}{4}\). Dấu "=" xảy ra khi \(\begin{cases}-\frac{1}{4}\le x\le-\frac{1}{3}\\x+\frac{1}{3}=0\end{cases}\)

\(\Leftrightarrow x=-\frac{1}{3}\)

Vậy Min D = \(\frac{1}{4}\Leftrightarrow x=-\frac{1}{3}\)

baì đâu vậy bạn

xin lỗi t ms học lớp 6-7

a) \(\left(2x-1\right)^2-4x^2+1=0\)

\(\Leftrightarrow\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x-1-2x-1\right)=0\)

\(\Leftrightarrow-2\left(2x-1\right)=0\)

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

b) \(3x\left(x+5\right)-2x-10=0\)

\(\Leftrightarrow3x\left(x+5\right)-2\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+5=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\x=\frac{2}{3}\end{array}\right.\)

c) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)

\(\Leftrightarrow\left[2\left(2x+7\right)+3\left(x+3\right)\right]\cdot\left[2\left(2x+7\right)-3\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(4x+14+3x+9\right)\left(4x+14-3x-9\right)=0\)

\(\Leftrightarrow\left(7x+23\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}7x+23=0\\x+5=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{23}{7}\\x=-5\end{array}\right.\)

a) Đăt \(x^2+x=t\) khi đó bt trở thành:

 \(t^2-2t-15=t^2+3t-5t-15=t\left(t+3\right)-5\left(t+3\right)\\ =\left(t+3\right)\left(1-5\right)=\left(x^2+x+3\right)\left(x^2+x-5\right)\)

C) \(\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)+15=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(x^2+8x+11=t\) thì bt trở thành

\(\left(t-4\right)\left(t+4\right)+15\)

\(=t^2-16+15=t^2-t-15t+15=t\left(t-1\right)-15\left(t-1\right)\)

\(=\left(t-1\right)\left(t-15\right)=\left(x^2+8x+11-1\right)\left(x^2+8x+11-15\right)\\ =\left(x^2+8x+10\right)\left(x^2+8x-4\right)\)

thanks nha nhưng cô giáo chữa cho mk rùi

\(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}\)

\(=\frac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)+2-3x}{2x\left(2x-1\right)}\)

\(=\frac{2x-1-6x^2+3x+6x^2-4x+2-3x}{2x\left(2x-1\right)}\)

\(=\frac{-2x+1}{2x\left(2x-1\right)}\)

\(=-\frac{1}{2x}\)

\(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}\)

\(=\frac{1-3x}{2x}+\frac{3x-2}{2x-1}-\frac{3x-2}{2x\left(2x-1\right)}\)

\(=\frac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)-\left(3x-2\right)}{2x\left(2x-1\right)}\)

\(=\frac{2x-1-6x^2+3x+6x^2-4x-3x+2}{2x\left(2x-1\right)}\)

\(=\frac{-2x+1}{2x\left(2x-1\right)}=\frac{-\left(2x-1\right)}{2x\left(2x-1\right)}=-\frac{1}{2x}\)

điều kiện x khác {0,1/2}

\(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}=\frac{2x-6x^2-1+3x+6x^2-4x-3x+2}{2x\left(2x-1\right)}\)

=\(\frac{-2x+1}{2x\left(2x-1\right)}=-\frac{1}{2x}\)

\(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)

\(=\frac{4\left(x-2\right)+2\left(x+2\right)-\left(5x-6\right)}{x^2-4}\)

\(=\frac{4x-8+2x+4-5x+6}{x^2-4}\)

\(=\frac{x+2}{\left(x-2\right)\left(x+2\right)}=\frac{1}{x-2}\)

điều kiện x khác {2,-2}

\(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}=\frac{4x-8+2x+4-5x+6}{x^2-4}\)

=\(\frac{x+2}{\left(x+2\right)\left(x-2\right)}=\frac{1}{x-2}\)

\(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)

\(=\frac{4}{x+2}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\)

\(=\frac{4\left(x-2\right)+2\left(x+2\right)+6-5x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x-8+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x+2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{1}{x-2}\)

\(2222^{5555}+5555^{2222}\)

\(=3^{5555}+4^{2222}\)

\(=243^{1111}+16^{1111}\)

\(=5^{1111}+2^{1111}\)

\(=\left(-2\right)^{1111}+2^{1111}\)

\(=0\left(mod7\right)\) (chia hết cho 7)